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3 years ago

What is radiation? Explain in your own words.

Physics

2 answers:

tatyana61 [14]3 years ago

4 0

This website explains: https://study.com/academy/lesson/what-is-radiation-lesson-for-kids-definition-examples.html

Evgesh-ka [11]3 years ago

3 0

Answer:

An energy source in which we use to power most powerplants and reactors.

Explanation:

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A coil of conducting wire carries a current of i(t) = 14.0 sin(1.15 ✕ 103t), where i is in amperes and t is in seconds. A second

Advocard [28]

Answer:

The peak emf in second coil is 1.876 V

Explanation:

Given :

Inductance $L = 130 \times 10^{-6}$ H

The current $I(t) = 14 \sin(1.15\times 10^{3} t)$

We compare above equation with standard equation,

$I(t) = I_{o} \sin (\omega t + \phi)$

From above equation we have,

$\omega = 10^{3}$ and $\phi = 1.15$

Find the inductive resistance,

$X_{L} = \omega L$

$X_{L} = 10^{3} \times 130 \times 10^{-6}$

$X_{L} = 0.134$

The peak emf in second coil is,

$V = I_{o} X_{L}$

$V = 14 \times 0.134$

$V = 1.876$ V

Therefore, the peak emf in second coil is 1.876 V

8 0

3 years ago

A 1400.0 kg car crests a 3200.0 m pass in the mountains and briefly comes to rest. The car descends 1000 m before climbing and c

rjkz [21]

Answer:

a) v = 88.54 m/s

b) vf = 26.4 m/s

Explanation:

Given that;

m = 1400.0 kg

by using the energy conservation

loss in potential energy is equal to gain in kinetic energy

mg × ( 3200-2800) = 1/2 ×m×v²

1400 × 9.8 × 400 = 0.5 × 1400 × v²

5488000 = 700v²

v² = 5488000 / 700

v² = 7840

v = √7840

v = 88.54 m/s

Work done by all forces is equal to change in KE

W_gravity + W_non - conservative = 1/2×m×(vf² - vi²)

we substitute

1400 × 9.8 × ( 3200-2800) - (5 × 10⁶) = 1/2 × 1400 × (vf² -0 )

488000 = 700 vf²

vf² = 488000 / 700

vf² = 697.1428

vf = √697.1428

vf = 26.4 m/s

4 0

3 years ago

Ammonia can be represented by the chemical formula NH3. Which of the following BEST describes ammonia

True [87]

The answer is B, a compound. :)

8 0

3 years ago

Read 2 more answers

Consider a particle of mass m which can move freely along the x axis from -a/2 to a/2, but which is strictly prohibited from bei

nalin [4]

Answer:

φ = B sin (2π n/a x)

Explanation:

In quantum mechanics when a particle moves freely it implies that the potential is zero (V = 0), so its wave function is

φ = A cos kx + B sin kx

we must place the boundary conditions to determine the value of the constants A and B.

In our case we are told that the particle cannot be outside the boundary given by x = ± a / 2

therefore we must make the cosine part zero, for this the constant A = 0, the wave function remains

φ = B sin kx

the wave vector is

k = 2π /λ

now let's adjust the period, in the border fi = 0 therefore the sine function must be zero

φ (a /2) = 0

0 = A sin (2π/λ a/2)

therefore the sine argument is

2π /λ a/2 = n π

λ= a / n

we substitute

φ = B sin (2π n/a x)

7 0

3 years ago

Doubly ionized lithium Li2+ (Z = 3) and triply ionized beryllium Be3+ (Z = 4) each emit a line spectrum. For a certain series of

EleoNora [17]

Answer:

tex]\lambda_{Be}[/tex] = 22.78 nm

Explanation:

Bohr's model for the hydrogen atom has been used by other atoms with a single electric charge by changing the number of charges by the charge of the new atom (atomic number)

$E_{n}$ = k e² / 2a₀ (1 /n²)

ao = h'² / k m e² h' = h/2πi

For another atom with a single electron in the last layer

a₀ ’= h’² / k m (Ze)²

a₀ ’= a₀ / Z²

Therefore, when replacing in the equation

$E_{n}$ = - Z² Eo/n²

E₀ = 13,606 eV

The transition occurs when the electron stops from one level to another

$E_{n}$ - $E_{m}$ = Z² E₀ (1 / n² - 1 / m²) = Z² ΔE

Let's relate this expression to the wavelength

c = λ f

E = h f

E = h c /λ

h c / λ = Z² ΔE

λ = 1 / Z² (hc / ΔE)

λ = 1 / Z² λ_hydrogen

Let's apply this last equation to our case

Lithium Z = 3

$E_{n}$ = - 9 Eo / n²

40.5 10-9 = 1/9 λ_hydrogen

Beryllium Z = 4

λ = 1/16 λ_hydrogen

Let's write our two equations is and solve

40.5 10-9 = 1/9 λ_hydrogen

tex]\lambda_{Be}[/tex] = 1/ 16 λ_hydrogen

40.5 10⁻⁹ = 1/9 (16 $\lambda_{Be}$ )

tex]\lambda_{Be}[/tex] = 40.5 9/16

tex]\lambda_{Be}[/tex] = 22.78 nm

6 0

3 years ago